A coil is connected to an AC source with peak | vedclass

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(D) Given: Peak emf, $V_0 = 8 \, V$, Frequency $f = \frac{30}{\pi} \, Hz$, Resistance $R = 8 \, \Omega$, Average power $P_{avg} = 0.4 \, W$. $V_{rms}

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$0.4$

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(D) Given: Peak emf, $V_0 = 8 \, V$, Frequency $f = \frac{30}{\pi} \, Hz$, Resistance $R = 8 \, \Omega$, Average power $P_{avg} = 0.4 \, W$. $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{8}{\sqrt{2}} \, V$. Average power is given by $P_{avg} = V_{rms} I_{rms} \cos \phi = V_{rms} \left(\frac{V_{rms}}{Z}ight) \left(\frac{R}{Z}ight) = \frac{V_{rms}^2 R}{Z^2}$. Substituting the values: $0.4 = \frac{(8/\sqrt{2})^2 	imes 8}{Z^2} = \frac{32 	imes 8}{Z^2} = \frac{256}{Z^2}$. $Z^2 = \frac{256}{0.4} = 640 \, \Omega^2$. Since $Z^2 = R^2 + X_L^2$, we have $640 = 8^2 + X_L^2 = 64 + X_L^2$. $X_L^2 = 640 - 64 = 576$, so $X_L = 24 \, \Omega$. Using $X_L = 2 \pi f L$, we get $24 = 2 \pi (\frac{30}{\pi}) L = 60 L$. $L = \frac{24}{60} = 0.4 \, H$.

$A$ coil is connected to an $AC$ source with peak emf, $8 \, V$ and frequency $\frac{30}{\pi} \, Hz$. The coil has a resistance of $8 \, \Omega$. If the average power dissipated by the coil is $0.4 \, W$, then the inductance of the coil is (in $, H$)

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